## integral (cos x)^2 dx

Do any one have an idea how to calculate integral of (cos x)^2 ? Or is it even possible? I tried some substitutions and/or rules of trigonometry, like cosxcosx+sinxsinx=1, but it didn't help. Thank you!
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 Recognitions: Science Advisor cos2x+sin2x=1 cos2x-sin2x=cos2x Therefore cos2x=(1+cos2x)/2 I'll let you finish.
 Thank you. :) integral (cos x)^2 dx

## integral (cos x)^2 dx

dont you have to use half angle identities to get integral of cos^2 ?
 Recognitions: Gold Member Science Advisor Staff Emeritus No, double angle formulas as mathman said.
 an easy way to remember the solution to this common integral, when integrating over a whole period: cos^2 x + sin ^2 x =1 $$\int cos^2 x = \int sin^2 x$$ , at least when you integrate over a whole period $$\int cos^2 x + \int sin^2 x =$$ length of a period so the integral gives length of a period divided by 2
 Why does this thread have over 16,000 views? edit: Oh, it's four years old.
 First use the half-angle formula to change the cos(x)^2 to (1+cos(2x))/2... This will allow you to break the integral into two seperate problems much easier to solve integral{ 1/2dx + integral{ cos(2x)dx Then you will have x/2 + (sin(2x)/2) + C
 What the, that's not even correct. If you're gonna revive a 5-year old thread, at least make sure you don't have arithmetic errors.
 sin(2x)/4 ;)
 use the euler's formula cos x= [e^ix+e^-ix ] [-------------] [ 2 ]
 http://www.5min.com/Video/An-Introdu...sine-169056088 Why doesn't the student, after nearly 6 years of unsuccessfully attempting this crazy integral, try a visual aid?
 Mentor Blog Entries: 10 This is crazy. The very first reply, post #2, answered the question. Six years ago!